Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $y \neq 0$. $x = \dfrac{21y + 70}{-2} \times \dfrac{7y}{8y(3y + 10)} $
Explanation: When multiplying fractions, we multiply the numerators and the denominators. $x = \dfrac{ (21y + 70) \times 7y } { -2 \times 8y(3y + 10) } $ $ x = \dfrac {7y \times 7(3y + 10)} {-2 \times 8y(3y + 10)} $ $ x = \dfrac{49y(3y + 10)}{-16y(3y + 10)} $ We can cancel the $3y + 10$ so long as $3y + 10 \neq 0$ Therefore $y \neq -\dfrac{10}{3}$ $x = \dfrac{49y \cancel{(3y + 10})}{-16y \cancel{(3y + 10)}} = -\dfrac{49y}{16y} = -\dfrac{49}{16} $